A 28.4378 g sample of impure magnesium carbonate was heated to complete decomposition
according to the equation
MgCO3(s) → MgO(s) + CO2(g).
After the reaction was complete, the solid
residue (consisting of MgO and the original
impurities) had a mass of 17.6846 g. Assuming that only the magnesium carbonate had
decomposed, how much magnesium carbonate was present in the original sample?
Answer in units of g
I'll give you the steps on how to solve:
(1) Get the molecular weights (MW) of MgCO3 and MgO.
-> I think you know how to do this one, you just go to periodic table and find the masses of the individual atoms and add them according to the chemical formulas.
(2) Let x = grams of PURE MgCO3 in the 28.4378
-> Therefore, we can say that 28.4378 - x = total or original impurities
(3) Using x, solve for the mass of MgO produced. Follow the formula:
-> x (1 mol MgCO3 / MW MgCO3) * (1 mol MgO / 1 mol MgCO3) * (MW MgO / 1 mol MgO) = (MW MgO / MW MgCO3) * x
(4) Then, add the mass of original impurity with the mass of MgO produced, which is equal to 17.6846
-> (28.4378 - x) + ((MW MgO / MW MgCO3) * x) = 17.6846
(5) Finally, solve for x.
Hope this helps~ :3
To find the mass of magnesium carbonate present in the original sample, we need to calculate the mass of the magnesium carbonate that decomposed during the reaction.
First, we calculate the mass of the magnesium carbonate that remained in the solid residue after the reaction:
Mass of solid residue = Mass of original impure magnesium carbonate - Mass of magnesium carbonate decomposed
Mass of solid residue = 17.6846 g - 0 g (assuming only magnesium carbonate decomposed)
Now, we subtract the mass of the solid residue from the total mass of the original impure magnesium carbonate:
Mass of magnesium carbonate = Mass of original impure magnesium carbonate - Mass of solid residue
Mass of magnesium carbonate = 28.4378 g - 17.6846 g
Mass of magnesium carbonate = 10.7532 g
Therefore, there was 10.7532 grams of magnesium carbonate present in the original sample.